# Finding path for maximum value of line integral

I need help in the following question:

I have a field: $$F=\left(y^3-3y+xy^2,3x-x^3+x^2y\right)$$ bounded in region $$D$$ defined by $$x^2+y^2\leq 2.$$ I need to find a path $$C$$ that goes from $$(1,1)$$ to $$(-1,-1)$$ inside $$D$$ such that the value of the line integral/work of $$F$$ on the path $$C$$ is maximum.

I have thought about the edge of the circle as a possible path because the magnitude of $$F$$ seems to grow with the distance from $$(0,0),$$ but the field is not tangential so I don't know how to prove it's actually the maximum. So I am out of ideas; thanks in advance.

• You might be able to adapt this idea (mathhelpboards.com/potw-university-students-34/…) for your problem. The issue, of course, is that the problem to which I linked had a closed curve, whereas yours is not. Green's Theorem only applies for a simple closed curve. Working on it... Jun 14 '19 at 16:42
• Hello sir, if it helps, in the previous section they asked to explain intuitively why for every path C1 that isn't the edge of the circle there is a path C2 such that the value of the line integral for C1 is smaller than C2. This pretty much screams that the edge of the circle is the maximum work path. I thought I'd start with this, but now it seems like the given order is the natural one.. Jun 14 '19 at 16:52

Let's examine a related problem first: finding a simple, closed curve $$C$$ from $$(1,1)$$ to $$(-1,-1)$$ and back again that maximizes $$\displaystyle\oint_C\mathbf{F}\cdot d\mathbf{r}.$$ Let $$G$$ be the region enclosed by $$C.$$ Using the notation $$\mathbf{F}=(F_x, F_y),$$ not to be confused with taking the partial derivatives, which I will make explicit, we have by Green's Theorem that \begin{align*} \oint_C \mathbf{F}\cdot d\mathbf{r}&=\iint_G\left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)dA \\ &=\iint_G\left[\left(3-3x^2+2xy\right)-\left(3y^2-3+2xy\right)\right]dx\,dy \\ &=\iint_G\left[6-3x^2-3y^2\right]dx\,dy. \end{align*} To maximize this integral, we want to pick up all the region where the integrand is positive. But this is all of $$D,$$ since \begin{align*} 6-3x^2-3y^2&\ge 0 \\ 6 &\ge 3\left(x^2+y^2\right)\\ 2&\ge x^2+y^2. \end{align*} This exactly describes $$D.$$ To pick up all of $$D,$$ we will just do the circle of radius $$\sqrt{2},$$ which is the boundary of $$D.$$ The integral will become \begin{align*} \iint_D\left[6-3x^2-3y^2\right]dx\,dy&=\int_0^{2\pi}\int_0^{\sqrt{2}}\left(6-3r^2\right)r\,dr\,d\theta \\ &=6\pi. \end{align*} But notice that the final integrand we had was $$\theta$$-independent. So from the area integral perspective, we wanted to pick up as much of $$D$$ as possible, and we needed to skirt around it on its boundary. Moreover, it didn't matter which direction we chose: we could have gone clockwise or counter-clockwise.
Now, to solve the original problem, we can see by angle symmetry that we just want to chop the region in half diagonally from $$(1,1)$$ to $$(-1,-1),$$ and go around the circle's edge. The maximum value will just be half of what we computed before, or $$3\pi.$$
• All you have to do is constrain the "back again" to be the line segment from $(-1,-1)$ to $(1,1)$, so that the region has that as its lower boundary. Note that the work along that line segment is $0$ anyhow. Jun 14 '19 at 17:04
• I think the bit about the direction needs to be clarified. The direction does matter because the contour in Green's theorem is assumed to be positively oriented. There is only one choice for $C$: the semicirle above the line $y = x$. Rephrasing your argument, the scalar curl is positive, therefore the integral over this $C$ minus the integral over any other path from $(1, 1)$ to $(-1, -1)$ is positive, because we get a closed contour traversed in the positive direction. Jun 14 '19 at 21:21